This code calculate the SUM of elements in each column of array T(5,3) and stored in array K(3). For I = 1 To 5 For J = 1 To 3 T(I, J) = InputBox("T") %3D %3D
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- in this javascript function ,how tochange the code to sort the array elements in descending order function get_odd_num(){ var max_val= 1000; var randNum= Math.random()*max_val; if(randNum %2 == 0){ if(randNum == max_val){ randNum = randNum -1 ; }else{ randNum = randNum +1 ; } }} function sortArray() { var array =[]; for (var i = 0; i < 10; i++) { array.push(get_odd_num()); } var temp = 0; for (var i = 0; i < array.length; i++) { for (var j = i; j < array.length; j++) { if (array[j] > array[i]) { temp = array[j]; array[j] = array[i]; array[i] = temp; } } } return array;} console.log(sortArray());Create a main() method in which we will create a 5 element array of RegularTriangle objects. For each element, create a random int between 1-10 and create the element instance using this number as side. Once created, calculate the average perimeter of these arrays. Hint: use the following code to generate random number import java.util.Random; Random rand = new Random(); int n = rand.nextInt(100); Add screenshot of outputThe editTemperatures() method takes in two parameters: integer array tempData and integer numMore. Complete editTemperatures() to create a new array called arrayCopy with the same elements as tempData, and the size increased by numMore. Ex: If the input is: 10 48 49 18 3 then the output is: 10 48 49 18 0 0 0 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 Scannerscnr=newScanner(System.in); int[] degreesFahrenheit=newint[4]; intinput; inti; for (i=0; i<degreesFahrenheit.length; ++i) { degreesFahrenheit[i] =scnr.nextInt(); } // Read number of values to increase array size by input=scnr.nextInt(); degreesFahrenheit=editTemperatures(degreesFahrenheit, input); for (i=0; i<degreesFahrenheit.length; ++i) { System.out.print(degreesFahrenheit[i] +" "); } } }
- # the method take another circular array and returns a linear array containing the common elements between the two circular arrays. def intersection(self, c2): # To Do print("==========Test 10==========") lin_arr9 = [10, 20, 30, 40, 50, None, None, None] c10 = CircularArray(lin_arr9, 5, 5) c10.printFullLinear() # This should print: 40, 50, None, None, None, 10, 20, 30 lin_arr10 = [5, 40, 15, 25, 10, 20, 5, None, None, None, None, None] c11 = CircularArray(lin_arr10, 8, 7) c11.printFullLinear() # This should print: 10, 20, 5, None, None, None, None, None, 5, 40, 15, 25 output = c10.intersection(c11) print(output) # This should print: [10, 20, 40]given an array of integer values , return true if 6 appears as either the first or last element in the array. firstLast6([1,2,36]) true firstLast([6,1,2,3]) true firstLast([13,6,1,2,3]) falseWrite a method that finds the largest element in an arrayof double values using the following header:public static double max(double[] array)Write a test program that prompts the user to enter ten numbers, invokes thismethod to return the maximum value, and displays the maximum value. Here isa sample run of the program: Enter ten numbers: 1.9 2.5 3.7 2 1.5 6 3 4 5 2 ↵EnterThe minimum number is: 6
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- An array having a character subscript can be indexed using a computer language, as in CHR_ARRAY['A':'D']. In this scenario, the characters' ordinal numbers (ORDs), such as ORD('A') = 1, ORD('B') = 2, ORD('Z') = 26 and so on, are used to designate the array's elements, CHR_ARRAY['A'], CHR_ARRAY['B'], and so on. Beginning at position 500, two arrays called TEMP[1:5, -1:2] and CODE['A':'Z', 1:2] are now kept in the RAM. Furthermore, CODE outperforms TEMP in storage. Determine the locations of the following addresses: (i) TEMP[5, -1], (ii) CODE['N',2], and (iii) CODE['Z',1].in java Integer numElements is read from input, representing the number of integers to be read next. Then, the remaining integers are read and stored into array valuesArray. For each element in valuesArray that is divisible by 9, assign the element with the element's current value plus 1. Ex: If the input is: 3 81 99 30 then the output is: Original values: 81 99 30 New values: 82 100 30 Note: (x % 9 == 0) returns true if x is divisible by 9. 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 for (i=0; i<valuesArray.length; ++i) { valuesArray[i] =scnr.nextInt(); } System.out.print("Original values: "); for (i=0; i<valuesArray.length; ++i) { System.out.print(valuesArray[i] +" "); } System.out.println(); /* Your code goes here */ System.out.print("New values: "); for (i=0; i<valuesArray.length; ++i) { System.out.print(valuesArray[i] +" "); } System.out.println(); } }In Java Write an application that inputs five numbers, each between 10 and 100, inclusive. As each number is read, display it only if it’s not a duplicate of a number already read. Provide for the “worst case,” in which all five numbers are different. Use the smallest possible array to solve this problem. Display the complete set of unique values input after the user enters each new value. Rough output in screenshot